answer:from collections import deque def shortestPath(grid, k): m, n = len(grid), len(grid[0]) visited = [[[False for _ in range(k + 1)] for _ in range(n)] for _ in range(m)] q = deque([(0, 0, 0, k)]) moves = [(0, 1), (0, -1), (1, 0), (-1, 0)] steps = 0 while q: for _ in range(len(q)): x, y, steps, k_remaining = q.popleft() if x == m - 1 and y == n - 1: return steps for dx, dy in moves: new_x, new_y, = x + dx, y + dy if 0 <= new_x < m and 0 <= new_y < n: new_k_remaining = k_remaining - grid[new_x][new_y] if new_k_remaining >= 0 and not visited[new_x][new_y][new_k_remaining]: visited[new_x][new_y][new_k_remaining] = True q.append((new_x, new_y, steps + 1, new_k_remaining)) steps += 1 return -1

answer:import heapq def dijkstra(graph, start, end): queue = [] heapq.heappush(queue, (0, start)) distances = {node: float('infinity') for node in graph} distances[start] = 0 shortest_path = {} while queue: current_distance, current_node = heapq.heappop(queue) if current_distance > distances[current_node]: continue for neighbor, weight in graph[current_node].items(): distance = current_distance + weight if distance < distances[neighbor]: distances[neighbor] = distance heapq.heappush(queue, (distance, neighbor)) shortest_path[neighbor] = current_node path = [] while end: path.append(end) end = shortest_path.get(end) return distances, path[::-1] graph = { 'A': {'B': 3, 'C': 4}, 'B': {'A': 3, 'D': 2}, 'C': {'A': 4, 'D': 6, 'E': 2}, 'D': {'B': 2, 'C': 6, 'E': 1}, 'E': {'C': 2, 'D': 1} } distances, path = dijkstra(graph, 'A', 'E') print("Shortest Path:", path) print("Shortest Distance:", distances['E'])

answer:```cpp #include <iostream> using namespace std; int factorial(int n); int main() { int num; cout << "Enter a positive integer: "; cin >> num; if (num >= 0) { cout << "Factorial of " << num << " = " << factorial(num); } else { cout << "Invalid input. Please enter a positive integer."; } return 0; } int factorial(int n) { if(n > 1) return n * factorial(n - 1); else return 1; } ``` This C++ program accepts an integer input and outputs its factorial using recursion. The `factorial` function is defined recursively, with the base case being when `n <= 1`, in which case it returns 1. Otherwise, it calls itself with `n-1` as the argument and multiplies the result by `n`.

answer:Let's break down this problem step-by-step: 1. First, let's calculate the probability of drawing a white ball first and then a black ball: - The probability of drawing a white ball first is 7/15 (since there are 7 white balls out of 15 total balls). - After drawing a white ball, there are now 14 balls left in the bag, including 6 black balls. So, the probability of drawing a black ball next is 6/14. - Therefore, the probability of drawing a white ball first and then a black ball is (7/15) * (6/14). 2. Next, let's calculate the probability of drawing a black ball first and then a white ball: - The probability of drawing a black ball first is 6/15 (since there are 6 black balls out of 15 total balls). - After drawing a black ball, there are now 14 balls left in the bag, including 7 white balls. So, the probability of drawing a white ball next is 7/14. - Therefore, the probability of drawing a black ball first and then a white ball is (6/15) * (7/14). 3. Since we want to find the probability of either drawing a white ball first and then a black ball or drawing a black ball first and then a white ball, we can add these two probabilities together: - P(one white ball and one black ball) = [(7/15) * (6/14)] + [(6/15) * (7/14)] The final answer is that the probability of drawing one white ball and one black ball is [(7/15) * (6/14)] + [(6/15) * (7/14)], which simplifies to 7/15 or approximately 46.67%.